There are several ways in which we could do this. One strategy might be to expand the brackets and compare the terms \[\textrm{Numerator} \ A: (10^a+1)(10^b+2)=10^{a+b}+2\times10^a + 10^b + 2\] and \[\textrm{Numerator} \ B: (10^b+1)(10^a+2)=10^{a+b}+2\times10^b+10^a+2.\]

Only the middle terms differ so if we consider this part of the numerator of \(A\): \[C: 2\times10^a+10^b\] and the equivalent part of the numerator of \(B\): \[D: 2\times10^b+10^a.\]

If \(a>b\) then in expression C the first term will dominate and in expression D the second term will dominate. Since the first term of C is double the second term in D, expression C must be greater.

Similarly, if \(b>a\) then in expression C the second term will dominate and in expression D the first term will dominate. Since the first term of D is double the second term of C expression D must be greater.

This means that overall if \(a>b\) then \[\textrm{Fraction} \ A > \textrm{Fraction} \ B\] and if \(b>a\) then \[\textrm{Fraction} \ B > \textrm{Fraction} \ A.\]

We could have reached this solution more quickly if we had noticed at the above expressions (showing the differing middle terms) can be reduced to \[C: 10^a \quad D: 10^b.\]

Thinking about this graphically we see that \(10^x\) is an increasing function and so fraction A will be greater than fraction B when \(a>b\).

We should check that this makes sense when we substitute values in place of \(a\) and \(b\). Substituting \(a=1\) and \(b=0\) into fractions A and B gives \[A: \frac{10^1+1}{10^1+2} = \frac{11}{12}\] and \[B: \frac{10^0+1}{10^0+2} = \frac{2}{3}.\]

Then we can show that \(\frac{11}{12}>\frac{2}{3}\) and our solution holds.

We could have approached this question slightly differently if we had started from the initial requirement that fraction A is greater than fraction B.